Transcript 4.1 Mathematical Expectation

Known Probability Distributions
• Engineers frequently work with data that can be
modeled as one of several known probability
distributions.
• Being able to model the data allows us to:
 model real systems
 design
 predict results
• Key discrete probability distributions include:
 binomial
 negative binomial
 hypergeometric
 Poisson
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Discrete Uniform Distribution
• Simplest of all discrete distributions
 All possible values of the random variable have the
same probability, i.e.,
f(x; k) = 1/ k, x = x1 , x2 , x3 , … , xk
• Expectations of the discrete uniform distribution
k

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k
 xi
i 1
k
and
2 
 ( x i   )2
i 1
k
2
Binomial & Multinomial Distributions
• Bernoulli Trials
 Inspect tires coming off the production line. Classify each as
defective or not defective. Define “success” as defective. If historical
data shows that 95% of all tires are defect-free, then P(“success”) =
0.05.
 Signals picked up at a communications site are either incoming
speech signals or “noise.” Define “success” as the presence of
speech. P(“success”) = P(“speech”)
 Administer a test drug to a group of patients with a specific
condition. P(“success”) = ___________
• Bernoulli Process




n repeated trials
the outcome may be classified as “success” or “failure”
the probability of success (p) is constant from trial to trial
repeated trials are independent.
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Binomial Distribution
• Example:
Historical data indicates that 10% of all bits
transmitted through a digital transmission channel are
received in error. Let X = the number of bits in error in
the next 4 bits transmitted. Assume that the
transmission trials are independent. What is the
probability that
 Exactly 2 of the bits are in error?
 At most 2 of the 4 bits are in error?
 more than 2 of the 4 bits are in error?
• The number of successes, X, in n Bernoulli trials
is called a binomial random variable.
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Binomial Distribution
• The probability distribution is called the binomial
distribution.
 n  x nx
 b(x; n, p) =   p q
x
, x = 0, 1, 2, …, n
where p = _________________
q = _________________
• For our example,
 b(x; n, p) = _________________
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For Our Example …
• What is the probability that exactly 2 of the bits
are in error?
• At most 2 of the 4 bits are in error?
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Your turn …
• What is the probability that more than 2 of the 4
bits are in error?
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Expectations of the Binomial Distribution
• The mean and variance of the binomial
distribution are given by
μ = np
σ2 = npq
• Suppose, in our example, we check the next 20
bits. What are the expected number of bits in
error? What is the standard deviation?
μ = ___________
σ2 = __________ ,
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σ = __________
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Another example
A worn machine tool produces 1% defective parts. If we
assume that parts produced are independent, what is the
mean number of defective parts that would be expected
if we inspect 25 parts?
What is the expected variance of the 25 parts?
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Helpful Hints …
• Sometimes it helps to draw a picture.
Suppose we inspect the next 5 parts …
P(at least 3) 
P(2 ≤ X ≤ 4) 
P(less than 4) 
• Appendix Table A.1 (pp. 661-666) lists Binomial
Probability Sums, ∑rx=0b(x; n, p)
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Your turn …
• Use Table A.1 to determine
1. b(x; 15, 0.4) , P(X ≤ 8) = ______________
2. b(x; 15, 0.4) , P(X < 8) = ______________
3. b(x; 12, 0.2) , P(2 ≤ X ≤ 5) = ___________
4. b(x; 4, 0.1) , P(X > 2) = ______________
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Multinomial Experiments
• What if there are more than 2 possible outcomes?
(e.g., acceptable, scrap, rework)
• That is, suppose we have:
 n independent trials
 k outcomes that are
 mutually exclusive (e.g., ♠, ♣, ♥, ♦)
 exhaustive (i.e., ∑all kpi = 1)
• Then
n

 x1 x2
 p1 p2 ... pkxk
f(x1, x2, …, xk; p1, p2, …, pk, n) = 
 x1, x2 ,..., xk 
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Example
• Look at problem 22, pg. 126
x1 = _______
p1 = _______
x2 = _______
p2 = _______
x3 = _______
p3 = _______
n = _____
f( __, __, __; ___, ___, ___, __) =_________________
= __________________________________
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Hypergeometric Distribution
• Example*:
Automobiles arrive in a dealership in lots of 10. Five out
of each 10 are inspected. For one lot, it is know that 2
out of 10 do not meet prescribed safety standards.
What is probability that at least 1 out of the 5 tested from
that lot will be found not meeting safety standards?
*from Complete Business Statistics, 4th ed (McGraw-Hill)
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• This example follows a hypergeometric distribution:
 A random sample of size n is selected without replacement from N
items.
 k of the N items may be classified as “successes” and N-k are
“failures.”
• The probability associated with getting x successes in the
sample (given k successes in the lot.)
 k  N  k 
 

x  n  x 

P ( X  x )  h( x; N, n, k ) 
N 
 
n 
Where,
k = number of “successes” = 2
N = the lot size = 10
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n = number in sample
=5
x = number found
= 1 or 2
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Hypergeometric Distribution
• In our example,
P ( X  x )  P ( X  1)  P ( X  2)
 2 10  2   2 10  2 
 
  

1  5  1   2  5  2 

h(1;10,5,2)  h(2;10,5,2) 

10 
10 
 
 
5 
5 
= _____________________________
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Expectations of the Hypergeometric
Distribution
• The mean and variance of the hypergeometric distribution
are given by
nk

N
N n
k
k
2
 
* n * (1  )
N 1
N
N
• What are the expected number of cars that fail inspection in
our example? What is the standard deviation?
μ = ___________
σ2 = __________ ,
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σ = __________
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Your turn …
A worn machine tool produced defective parts for a period of
time before the problem was discovered. Normal sampling of
each lot of 20 parts involves testing 6 parts and rejecting the
lot if 2 or more are defective. If a lot from the worn tool
contains 3 defective parts:
1. What is the expected number of defective parts in a sample of
six from the lot?
2. What is the expected variance?
3. What is the probability that the lot will be rejected?
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Binomial Approximation
• Note, if N >> n, then we can approximate this with the
binomial distribution. For example:
Automobiles arrive in a dealership in lots of 100. 5 out of
each 100 are inspected. 2 /10 (p=0.2) are indeed below
safety standards.
What is probability that at least 1 out of 5 will be found
not meeting safety standards?
• Recall: P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
Hypergeometric distribution
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Binomial distribution
(Compare to example 5.14, pg. 129)
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Negative Binomial Distribution
• Example:
Historical data indicates that 30% of all bits
transmitted through a digital transmission channel are
received in error. An engineer is running an
experiment to try to classify these errors, and will start
by gathering data on the first 10 errors encountered.
What is the probability that the 10th error will occur on
the 25th trial?
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• This example follows a negative binomial distribution:
 Repeated independent trials.
 Probability of success = p and probability of failure = q = 1-p.
 Random variable, X, is the number of the trial on which the kth
success occurs.
• The probability associated with the kth success occurring on
trial x is given by,
 x  1 k x k
 p q , x  k, k  1, k  2,...
b * ( x; k, p )  
 k  1
Where,
k = “success number” = 10
x = trial number on which k occurs = 25
p = probability of success (error) = 0.3
q = 1 – p = 0.7
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Negative Binomial Distribution
• In our example,
 25  1
(0.3)10 (0.7)25 10
b * (25;10,0.3)  
10  1
= _____________________________
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Geometric Distribution
• Example:
In our example, what is the probability that the 1st bit
received in error will occur on the 5th trial?
• This is an example of the geometric distribution,
which is a special case of the negative binomial
in which k = 1.
 The probability associated with the 1st success
occurring on trial x is given by
g ( x; p)  pq x 1
= __________________________________
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Your turn …
A worn machine tool produces 1% defective parts. If we
assume that parts produced are independent:
1.What is the probability that the 2nd defective part will be the
6th one produced?
2.What is the probability that the 1st defective part will be seen
before 3 are produced?
3.How many parts can we expect to produce before we see the
1st defective part? (Hint: see Theorem 5.4, pg. 135)
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Poisson Process
• The number of occurrences in a given interval or
region with the following properties:
 “memoryless”
 P(occurrence) during a very short interval or small region
is proportional to the size of the interval and doesn’t
depend on number occurring outside the region or
interval.
 P(X>1) in a very short interval is negligible
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Poisson Process
• Examples:
 Number of bits transmitted per minute.
 Number of calls to customer service in an hour.
 Number of bacteria in a given sample.
 Number of hurricanes per year in a given region.
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Poisson Process
• Example
An average of 2.7 service calls per minute are received
at a particular maintenance center. The calls
correspond to a Poisson process. To determine
personnel and equipment needs to maintain a desired
level of service, the plant manager needs to be able to
determine the probabilities associated with numbers of
service calls.
What is the probability that fewer than 2 calls will be
received in any given minute?
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Poisson Distribution
• The probability associated with the number of
occurrences in a given period of time is given by,
e  t (  t ) x
p( x; t ) 
, x  0,1,2,...
x!
Where,
λ = average number of outcomes per unit time or region
= 2.7
t = time interval or region = 1 minute
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Our Example
• The probability that fewer than 2 calls will be
received in any given minute is …
P(X < 2) = P(X = 0) + P(X = 1)
= __________________________
• The mean and variance are both λt, so
μ = _____________________
• Note: Table A.2, pp. 667-669, gives Σt p(x;μ)
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Poisson Distribution
• If more than 6 calls are received in a 3-minute
period, an extra service technician will be
needed to maintain the desired level of service.
What is the probability of that happening?
μ = λt = _____________________
P(X > 6) = 1 – P(X < 6)
= _____________________
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Poisson Distribution
50
Frequency
40
30
20
10
0
Calls per minute
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Poisson Distribution
The effect of λ on the Poisson distribution
0.4
0.35
0.3
0.25
0.2
0.15
0.1
1
2
3
4
5
6
7
8
9
0.05
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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